1997 AIME Problems/Problem 1

Problem

How many of the integers between 1 and 1000, inclusive, can be expressed as the difference of the squares of two nonnegative integers?

Solution

If we let the two squares be $a^2 - b^2 = x$, then by difference of squares we have $(a-b)(a+b) = x$. Notice that $a-b$ and $a+b$ have the same parities. This eliminates all numbers in the form of $4n+2$: when $x=2(2n+1)$ is factored, one of the factors must be even, but not both, so its factors cannot have the same parity.

The remaining $\boxed{750}$ numbers, we can describe specific squares which fit the conditions:

  • For all odd $x = 2n+1$, $(n+1)^2 - (n^2) = x$.
  • For all $x = 4n$, $(n+1)^2 - (n-1)^2 = x$.

See also

1997 AIME (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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